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Bernstein polynomials converge uniformly to every continuous function on [0,1][0,1]

Statement

If f:[0,1]Rf:[0,1]\to\mathbb R is continuous, then Bn(f)fB_n(f)\to f uniformly on [0,1][0,1].

Facts & Assumptions

Given: A continuous function f:[0,1]Rf:[0,1]\to\mathbb R and ε>0\varepsilon>0.

[L1]

A continuous function on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L2]

The Bernstein basis has the zeroth and centred second moment identities (The zeroth, first, and second centred moments of the Bernstein basis).

[L3]

For each positive real η\eta there is a natural N1N\ge1 with 1/N<η1/N<\eta (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon).

Proof

technique · direct
1.1

Choose δ>0\delta>0 such that f(s)f(x)<ε/2|f(s)-f(x)|<\varepsilon/2 whenever sx<δ|s-x|<\delta, and choose MM with fM|f|\le M.

L1choose
1.2

On the far part, (k/nx)2δ2(k/n-x)^2\ge\delta^2; hence its total basis weight is at most x(1x)/(nδ2)1/(4nδ2)x(1-x)/(n\delta^2)\le1/(4n\delta^2) by [L2].

L2algebra
2.1

Split the Bernstein sum into k/nx<δ|k/n-x|<\delta and its complement. The near part is at most ε/2\varepsilon/2 by the zeroth identity.

step 1.1L2algebra
2.2

Choose nn so large that 2M/(4nδ2)<ε/22M/(4n\delta^2)<\varepsilon/2. The far part is then below ε/2\varepsilon/2, uniformly in xx.

step 1.2L3algebra
3.1

The near and far estimates give Bn(f)(x)f(x)<ε|B_n(f)(x)-f(x)|<\varepsilon for every xx and all sufficiently large nn.

step 2.1step 2.2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources